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Properties of Gases and Gas Laws: Pressure, Temperature, Volume, and More

Learn about the properties of gases such as pressure, temperature, volume, density, and more. Discover the gas laws including Boyle's Law, Charles's Law, and Dalton's Law.

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Properties of Gases and Gas Laws: Pressure, Temperature, Volume, and More

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  1. Gases Gases A gas expand to occupy the entire volume it is placed in. Molecules in a gas translate freely between collisions, and they all behave alike regardless of their type. What are some of the properties of gases? Pressure, temperature, heat capacity, volume, density, molar volume, color, average speed of molecules, solubility (in water or other liquid), absorption, compressibility, gas-liquid equilibrium, composition, identity (compound or element), chemical properties, combustibility, stability Which of these properties are intensive properties, and which are extensive properties?

  2. Gases

  3. Gases Pressure Pressure: force per area (1 Pascal = 1 N m–2) Liquid pressure (explain these formulation in terms of physics) F W g * m g * V * d g * h * A * d P = --- = ---- = ---------- = --------------- = ----------------- = g * h * dA A A A A These equivalences are useful for unit conversions: 1 atm = 101.325 kPa = 76 cm Hg = 760 mm Hg (torr in honor of Torricelli) = 1.01325 b = 1013.25 mb (bar & m bar) = 14.6960 pounds / sq. inch

  4. Gases Torricelli’s Barometer Barometric pressure Explain Torricelli’s work(1608 - 1647) P = g h d 1 atm = 0.76 m Hg 13594 kg1 m3 Hg 9.80665 N1 kg = 101317 N m–2 (Pascal) = 101.32 k Pa

  5. Gases Torricelli Mercury Barometer Evangelista Torricelli invented the Torricelli Mercury Barometer in 1644. He used a long glass tube, closed at the upper end, open at the lower and filled with mercury.

  6. Gases No water!! Pump Water from a Well 1 0. 3 3m Water The specific gravity of mercury is 13.5939. If water is used for a barometer, what is the height of water corresponding to 1.00 atm? Solution: 76 cm Hg 13.5939 g1 cm3 Hg 1 cm3 H2O 1 g = 1033.14 cm H2O = 10.33 m H2O Explain water pump and depth of well Water water everywhere!What about a diver under water? Be sure to get that during lecture.

  7. Gases Pressure of Skater A skater weighing 70 kg stands on one foot and the contact between the blade and ice is 1 cm2.What is the pressure in torr sustained by the ice? Solution: 70 kg1 cm2 1 m3 Hg 13594 kg 1000 mm1 m 10000 cm21 m2 = 51493 mm Hg = 51493 torr = 67.8 atm

  8. Gases Avogadro’s Law The ABCD of gas laws are Avogadro’s, Boyle’s, Charle’s & Dalton’s laws of gases. Avogadro’s hypothesis: proposed in 1811at the same temperature T and pressure P, equal volumes contain equal amounts of gases in moles n. at the same temperature T and pressure P, the volume V of a gas is proportional to the number of molecules or number of moles n. VP,T = k n (k, a constant, 22.4 L at STP) Explain Avogadro’s hypothesis and implications Avogadro’s scientific contributions to science will be given. Explain Avogadro’s law in your language

  9. Gases Boyle’s Law Robert Boyle, (1621-91) Mathematical aspects of PV product will be discussed For a certain amount (constant n) of gas at constant temperature T, the volume V times the pressure P is a constant. P V n, T = constant = P1 V1 = P2 V2 How do you graph the Boyles law? State the law in another way. What curves are P-V plots? P T2n1 • P1 V1 T1n1 • P2 V2 V

  10. Gases Charle’s Law(law of Charles-Gay-Lussac) For a certain amount of gas at constant pressure, its volume, V, is directly proportional to its temperature T in Kelvin. Vn, P = b T (b is a constant)or Pn, V = b T (b is a constant) State the Charle’s law in another wayV1V2 ----- = ----- = b, V1T2 = V2 T1T1T2 Vn, Por Pn, V T Charles, Jacques-Alexandre-César (1746-1823, top) first to ascend in a H2-balloon, developed this law in 1787. Later Joseph Louis Gay-Lussac (lower) published a paper citing Charles’s law.

  11. Gases General Gas Equation Combining ABC gas laws, we have P1 V1T1 P2 V2T2 = = n R Subscripts 1 and 2 refer to different conditions for the same quantities of gases (n). Experiments show that one mole of gas at STP occupies 22.4 L.

  12. Gases The Ideal Gas Equation The ABC laws of gases can be combined into one and the result is an ideal gas equation. A+B+Cideal P V = n R T 1 atm * 22.4140 L R = ---------------------------- = 0.082057 L atm mol–1 K–1 1 mol 273.15 K 101.325 kPa * 22.414 L R = ----------------------------------- = 8.3145 L kPa mol–1 K–1 1 mol 273.15 K Please confirm that 1 kPa L = 1 J (1 L = 1e-3 m3)

  13. Gases Dalton’s law The Partial pressurePi is the pressure of a component in a mixture as if others don’t exist in that system – due to the fact all gases behave as if they are independent of each other. Pi = ni Dalton’s law: The total pressure Ptotal, of a mixture of gases is the sum of the partial pressures of the components. Ptotal = P1 + P2 + … + Pn = (n1 + n2 + … + nn) (Vis common to all) R TV correction R TV

  14. Gases Application of the Ideal gas Law Parameters of the ideal gas law: P, V, T, n and a constant R ideal P V = n R T Ideal gas law A+B+C At constant n and T, P1V1 = P2V2 Boyles law At constant n and VP= (n R / V) TP1 / T1 = P2 / T2 Charles law At constant n and PV = (n R / P) TV1 / T1 = V2 / T2ditto At constant P and TV = (R T / P) nV = k n Avogardro’s law

  15. Gases Gas Densities Evaluate the density of O2 (molar mass M = 32.0) at 300 K and 2.34 atm. Hint: Density d of a gas with mass m (= n M) and volume V m n M n dd = ----- = ------ ---- = ------V V V M Thus n R T d R TP = ------------- = -----------V M d = P M / R T Find relationship between density d and M. Manipulate symbols to get a useful formula before you calculate the quantities. Include and work out the units plsed=2.34 *32.0 / (0.08205*300) = ___

  16. Gases Reactions Involving Gases How much NaN3 is required to produce 12.0 L N2 gas at 302 K and 1.23 atm for the air bag in your designed Autotie? Solution:Equations: 2 NaN3 = 2 Na + 3 N2; n = V (P / R T ) N2 1.23 atm mol K 0.08205 L atm * 302 K (23+3*14) g NaN31 mol NaN3 2 mol NaN33 mol N2 12.0 L = ___ Work out N2 volume for 51 g NaN3 used under the same condition.

  17. Gases - reaction involving gases NO is made from oxidizing NH3 at 1123 K with platinum as a catalyst. How many liter of O2 at 300 K and 1 atm is required for each liter of NO measured also at 300 K and 1 atm? Solution: The reaction is: 4 NH3 + 5 O24 NO + 6 H2OSince n = ( P/RT ) V molar relationships are the same as volumetric relationship, providing T and P are the same. 1 L NO 5 mol O24 mol NO Complicated problem may have a simple solution, = 1.25 L O2 Further oxidation of NO leads to NO2, which is used to make HNO3, a valuable commodity. How much H2O is produced?

  18. Gases A Mixture of Gases What is the pressure exerted by the gas when 1.0 g of H2, 2.0 g of O2, and 0.1 g of CO2 are all confined in a 10.0 L cylinder at 321 K? Solution: Dalton’s law : ntotal = i ni (count molecules non-discriminately) P = n R T / V; 1 mol2 g H2 1 mol32 g O2 1 mol44 g CO2 n = = 0.585 mol 1.0 g H2 + 2.0 g O2 + 0.1 g CO2 8.314J * 321 K 10 L mol K P = 0.585 mol = 126 kPa (note 1 J = 1 kPa L) Calculate partial pressures of each gas plse

  19. Gases Vapor pressure The (saturated) vapor pressure is the partial pressure that is at equilibrium with another phase. Vapor pressure of ice Vapor pressure of water ExplainStructure of water moleculeHydrogen bondingStructure of waterStructure of iceVapor pressure of ice and waterRelative and absolute humidity

  20. Gases Collecting Gas Over Water When 1.234 g of a sample containing Ag2O is heated, 40.6 mL of O2 is collected over water at 296 K, and the atmosphere is 751 mmHg. Vapor pressure of water at 296 K is 21.1 mmHg. What is the percentage of Ag2O in the sample? Solution: Ag2O = 2 Ag + 0.5 O2n = P V / R T; R = 0.08205 L atm mol-1 K-1 P = (751 – 21.1) mmHg / 760 mmHg = 0.961 atm (PO2 = Ptotal – Pwater) V = 40.6 mL = 0.0406 L Mass of Ag2O = 231.7 g Ag2O1 mol Ag2O 0.961 atm*0.0406 L O20.08205 L atm mol-1 K-1*296 K 1 mol Ag2O0.5 mol O2 = 0.744 g Ag2O Percentage of Ag2O = 0.744 g Ag2O / 1.234 g = 0.603 Ag2O = 60.3 % Ag2O

  21. Gases Assumptions of Kinetic-molecular Theory 1. Gas is composed of tiny, discrete particles (molecules or atoms). 2. Particles are small and far apart in comparison to their own size. 3. Ideal gas particles are dimensionless points occupying zero volume. 4. Particles are in rapid, random, constant straight line motion. 5. There is no attractive force between gas molecules and between molecules and the sides of the container. 6. Molecules collide with one another and the sides of the container. 7. Energy is conserved but transferred in these collisions. 8. Energy is distributed among the molecules in a particular fashion known as the Maxwell-Boltzmann Distribution.

  22. Gases Kinetic-molecular Theory of Gases For N gas molecules, molecule mass = m, molecular mass = Mspeed = u, average speed = u , Avogadro’s number Nvolume = V, temperature = T, Pressure = P, Kinetic energy = ½ m u 2 Collision frequencyu N / V Pressure  (m v) (u) (N / V)  (N / V) m u2= 1/3 (N / V) m u2(1/3 due to 3-Dimensional space) P V = 1/3N m u2 = n R T (Meaning of T) Thus, 3 R T = NAm u2 correction Furthermore, u2 = 3 R T / M (Temperature and speed) Explain the significances of and apply these formulas for sciences

  23. Gases Molecular Speeds Distributions of speed of various gases will be demonstrated using a simulation program, and for each gas, three speeds are indicated. In the following: m = mass of a molecule, M = molar mass,R = gas constant, and k = R / Navogadro = Boltzmann constant. The most probable speed ump = (2 k T / m)1/2 = (2 R T / M)1/2 The root-mean-square speed urms = ( 3 k T / m)1/2 = (3 RT / M)1/2 The mean or average speeduave = ( 8 k T /π m)1/2 = (8 RT / π M)1/2

  24. Gases Diffusion of Gases All gases move together because they are subjected to the same pressure head. Different gases diffuse at different rates Diffusion contributes to net movement of O2 and CO2 across the alveolar-capillary membrane (breathe). Constant molecular motion. Diffusion from higher to lower concentration regions. Since uave = ( 8 k T /π m)1/2 = (8 RT / π M)1/2, (slide 24) 1Diffusion rate -------- Graham’s law M Discuss diffusion rates of H2, He, CH4, N2, O2, CO2, 235UF6, 238UF6at lecture

  25. Gases Diffusion problems Problems are usually to compare diffusion or effusion rates in the following terms: urms = ( 3 k T / m)1/2 = uave = ( 8 k T /π m)1/2 = diffusion rateeffusion time for same amount  distance traveled by molecules in certain periodamount of gas effused  3RTM  8RT πM 1M M

  26. Gases Comparing Effusion Rates If 1e20 N2 molecules effuse from an orifice in 1.0 min, how many H2 molecules will effuse the same orifice at the same condition (T P)? How many minutes will be required for the same number of H2 molecules to effuse? H2 effusion rate = MN2 / MH2 (N2effusion rate) = (28 / 2 )1e20 molecules/min) = __________ figure out value and units Time for 1e20 H2 molecules to effuse = (2 / 28 ) 1.0 min = __________ What is the effusion rate ratio of N2 and H2 or any two gases?

  27. Gases O2 P2 CO2 A P1 T Effusion and Life Breathing chemistry is complicated, and we can only scratches the surface!

  28. Gases The van der Waals equation for real gases Gases tend to behave ideally at high T and low P. Required T and P for ideality depends on gas properties and molar mass, and van der Waals proposed correction terms for the ideal gas equation for real gases. (P + ) (V – n b) = nR T Correction for intermolecular forces Correction for volume of molecules n 2aV 2 where a and b are gas-dependant constants. Gas aL2 atm mol-2b L mol-1 He 0.0341 0.0237 Ne 0.211 0.0171 N2 1.39 0.0391 O2 1.36 0.0318 CO2 3.59 0.0427 Cl2 6.49 0.0562 Explain the meaning of vdW eqn Note units for a and b

  29. Gases Application of van der Waal’s Equation What is the pressure of Cl2 at 300 K occupying 20.0 L according to vdW and ideal gas laws? Solution: Look up data for Cl2, a = 6.49 L2 atm mol-2, b = 0.0562 L mol-1,n = 1 mol, R = 0.08205 L atm K-1 mol-1 n R TV – n b n 2a V2 P = - 12 mol2 * 6.49 L2 atm mol-220.02 L2 1 mol * 0.08205 L atm mol-1 K-1*300 K20.0 L – 1 mol 0.0562 mol-1 L = - = _____________________ Please calculate the results, and P from the ideal gas law.

  30. Gases Problems related to van der Waal’s Equation What is the molar volume of Cl2 at 300 K and 1 atm according to vdW? Solution: Look up data for Cl2, a = 6.49 L2 atm mol-2, b = 0.0562 L mol-1,n = 1 mol, R = 0.08205 L atm K-1 mol-1 n R TP + n 2a / V2 V = + n b = 24.615 / (1+0.013) = 24.299 L try V = 22 L= 24 Ltry V = 24.434 Land calculate V = 24.625 / (1+ 0.011) = 24.434 L = ? Calculate P for a definite volume is easier, and using the successive method for V is interesting, but it’s a challenge. Find out how engineers deal with real gases.

  31. Gases Volume of vdW eqn What is the volume of occupied by 132 g CO2 gas at 12.5 atm and 300 K? Solution:Solve volume of van der Waals equation for V R T + b PP n2 aP n2 a bP V 3 – n ( ) V 2 + ( ) V – ( ) = 0 Derive please a = 3.59 L2 atm mol-2, b = 0.0427 L mol-1,n = 1 mol, R = 0.08205 L atm K-1 mol-1 This is similar to problem 6 –106, a question for practicing the successive approximation method.

  32. Gases Successive Method again What is the volume of occupied by 132 g CO2 gas at 12.5 atm and 300 K? Solution:a = 3.59 L2 atm mol-2, b = 0.0427 L mol-1, T = 300 Kn = 132/44 = 3 mol, R = 0.08205 L atm K-1 mol-1 n R TP + n 2a / V2 V = + n b = 3*0.08205*300 / (12.5 + 32* 3.59 / 12) + 3*.0427 = 1.78 L= 73.845 / (12.5 + 32*3.59 / 32) + 0.128 = 4.7 L= 73.845 / (12.5 + 32*3.59 / 42) + 0.128 = 5.2 L= 73.845 / (12.5 + 32*3.59 / 52) + 0.128 = 5.5 L= 73.845 / (12.5 + 32*3.59 / 5.62) + 0.128 = 5.58 L

  33. Gases Molecular Formula of Gas Combustion of 1.110 g hydrocarbon produces 3.613 CO2 and 1.109 g H2O. A 0.288 g sample of the same has a volume of 131 mL at 298 K and 753 mmHg. Find the molecular formula. Solution: C : H = : = 0.0821 : 0.123 = 1 : 1.5 = 2 : 3 3.61344 1.109*218 Empirical formula is C2H3 d R TP (0.288 g / 0.131 L–1) * 0.08205 L atm mol–1 K –1 * 298 K(753 / 760) atm M = = = 54.3_______ work out units C4H6 has a molar mass of 54.3. Confirm and conclusion please! A 2-step problem, similar to one in Advanced Exercises

  34. Gases Concepts to review – cont. Apply ideal gas low to various problems Calculate stoichiometric quantities using on gas law and reaction equations. Apply Dalton’s partial pressure equation Compare effusion or diffusion rates of gases

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